bellatrix wrote:
The quickest way to solve this problem is to plug in the options
area of photograph (inner rectangle) = 8*10 = 80
area of the border (outer - inner) = 144
area of border = area of outer rectangle - area of inner rectangle
plugging in option A as possible width of the border, the dimensions of the outer rectangle become, 8+3+3= 14 and 10+3+3=16 (equal width on either side)
area of outer rectangle = 14*16=224
area of border = 224-80=144 which satisfies the question
Answer is A.
I think algebra is equally if not more faster than plugging
dimensions of the photo are 8inch by 10inch , so area 80 sq in
let the border width be x
so outer area = ( 10 + 2x) * (8 + 2x)
given that area of border = 144 = outer area - inner area = ( 10 + 2x) * (8 + 2x) - 80
144= 80 + 36x + 4 x^2 -80
4x^2 + 36x - 144 = 0
x^2 + 9x -36 = 0
(x+12)(x-3)=0
x = -12 or 3
width cant be negetive so answer is 3 (option A)